\(ab+bc+ca=0\)

=>   \(\frac{ab+bc+ca}{abc}=0\)

=>  \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

Đặt:  \(\frac{1}{a}=x;\)\(\frac{1}{b}=y;\)\(\frac{1}{c}=z\)

Ta có:   \(x+y+z=0\)

=>  \(x^3+y^3+z^3=3xyz\)  (tự c/m, ko c/m đc ib)

hay  \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

\(B=\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}=abc.\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)

     \(=abc.\frac{3}{abc}=3\)

thanks

c: Ta có: \(a\left(a+2b\right)^3-b\left(2a+b\right)^3\)

\(=a^4+6a^3b+12a^2b^2+8ab^3-8a^3b-12a^2b^2-6ab^3-b^4\)

\(=a^4-2a^3b+2ab^3-b^4\)

\(=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)-2ab\left(a^2-b^2\right)\)

\(=\left(a-b\right)^3\cdot\left(a+b\right)\)

\(a^2+b^2+c^2-ab-ac-bc=0\\\Leftrightarrow 2a^2+2b^2+2c^2-2ab-2ac-2bc=0\\\Leftrightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)=0\\\Leftrightarrow (a-b)^2+(b-c)^2+(a-c)^2=0\)

Ta thấy: \(\left(a-b\right)^2\ge0\forall a;b\)

              \(\left(b-c\right)^2\ge0\forall b;c\)

              \(\left(a-c\right)^2\ge0\forall a;c\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\forall a;b;c\)

Mặt khác: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

nên: \(\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\a=c\end{matrix}\right.\)

\(\Leftrightarrow a=b=c\left(dpcm\right)\)

#\(Toru\)

\(\left(a+b+c\right)^2=a^2+b^2+c^2\Leftrightarrow ab+bc+ca=0\)

\(\Rightarrow a^3b^3+b^3c^3+c^3a^3=3a^2b^2c^2\)

Ta có:

\(\dfrac{bc}{a^2}+\dfrac{ac}{b^2}+\dfrac{ab}{c^2}=\dfrac{a^3b^3+b^3c^3+c^3a^3}{a^2b^2c^2}=\dfrac{3a^2b^2c^2}{a^2b^2c^2}=3\)

Kẻ đường cao BD ứng với AC. Do góc A tù \(\Rightarrow\) D nằm ngoài đoạn thẳng AC hay \(CD=AD+AC\) và \(\widehat{DAB}=180^0-120^0=60^0\)

Áp dụng định lý Pitago:

\(AB^2=BD^2+AD^2\) \(\Rightarrow BD^2=AB^2-AD^2\)

Trong tam giác vuông ABD:

\(cos\widehat{BAD}=\dfrac{AD}{AB}\Rightarrow\dfrac{AD}{AB}=cos60^0=\dfrac{1}{2}\Rightarrow AD=\dfrac{1}{2}AB\)

\(\Rightarrow BD^2=AB^2-\left(\dfrac{1}{2}AB^2\right)=\dfrac{3}{4}AB^2\)

Pitago tam giác BCD:

\(BC^2=BD^2+CD^2=\dfrac{3}{4}AB^2+\left(AD+AC\right)^2\)

\(=\dfrac{3}{4}AB^2+\left(\dfrac{1}{2}AB+AC\right)^2\)

\(=\dfrac{3}{4}AB^2+\dfrac{1}{4}AB^2+AB.AC+AC^2\)

\(=AB^2+AB.AC+AC^2\)

Hay \(a^2=b^2+c^2+bc\)

Áp dụng BĐT AM-GM ta có:

\(VT=\dfrac{c+ab}{a+b}+\dfrac{a+bc}{b+c}+\dfrac{b+ac}{a+c}\)

\(=\dfrac{c\left(a+b+c\right)+ab}{a+b}+\dfrac{a\left(a+b+c\right)+bc}{b+c}+\dfrac{b\left(a+b+c\right)+ac}{a+c}\)

\(=\dfrac{ac+bc+c^2+ab}{a+b}+\dfrac{a^2+ab+ac+bc}{b+c}+\dfrac{ab+b^2+bc+ac}{a+c}\)

\(=\dfrac{\left(b+c\right)\left(c+a\right)}{a+b}+\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}\)

\(\ge2\left(a+b+c\right)=2\left(a+b+c=1\right)\)

Khi \(a=b=c=\dfrac{1}{3}\)